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WARNING: SLIGHT NSFW

http://www.smbc-comics.com/index.php?db=comics&id=2934#comic

Uhh...guys, mathematically speaking, how accurate is this comic. From what I remember in High School

$$a^b= \underbrace{\,a\cdot a\cdot \ldots \cdot a \,}_{b\ \text{ times}} $$ So does that mean that $$i^i= \underbrace{\,i\cdot i\cdot \ldots \cdot i \,}_{i\ \text{ times}} $$

How can you multiply a number by itself an imaginary number of times?

Crostul
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Predestination
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    No. Definition of complex exponentiation is not so intuitive. – Crostul Aug 23 '15 at 13:14
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    The comic is absolutely accurate (mathematically). – dbanet Aug 23 '15 at 13:16
  • Thanks to @dbanet for clearing this up. Can we come up with some real world analogy to explain this? – Predestination Aug 23 '15 at 13:19
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    The one thing missing from the comic is that complex exponentiation is really a multivalued function – Ben Grossmann Aug 23 '15 at 13:19
  • For real search experts: find the other questions here about $i^i$ ... – GEdgar Aug 23 '15 at 13:20
  • @Predestination what do you think of the situation where $b$ is a rational number? What about if $b$ is a real number? – Ben Grossmann Aug 23 '15 at 13:20
  • @Omnomnomnom really? How is that? – dbanet Aug 23 '15 at 13:22
  • Good point @Omnomnomnom. As far as I know, when b was a rational number, we could write b in the form x/y. This would mean raising a to the power x and then finding the yth root of a. But I cannot imagine what we would do if b was irrational. Hmmm...the rabbit hole goes deeper than I originally thought. – Predestination Aug 23 '15 at 13:23
  • Consider $e^{i\pi(2k + 1/2)}$ for any integer $k$ – PM 2Ring Aug 23 '15 at 13:23
  • @PM2Ring on which planet is this called "multivalued"? – dbanet Aug 23 '15 at 13:24
  • @dbanet: Did I say it was multivalued? :) – PM 2Ring Aug 23 '15 at 13:26
  • @PM2Ring you didn't. But what's the purpose of your comment then? – dbanet Aug 23 '15 at 13:28
  • @dbanet: The purpose of my initial comment is to point out that the value of $i^i$ isn't unambiguously equal to $e^{-\pi/2}$, since the complex logarithm function is multivalued. But I certainly wasn't trying to imply that $e^z$ for complex $z$ is multivalued. – PM 2Ring Aug 23 '15 at 13:36
  • @PM2Ring ah, that. Well, there is no reason to continue to consider the complex logarithm function $\log$ multivalued. The only way to consistently extend real exponentiation to complex exponentiation is to define $a^b=e^{b\log(a)}$ where $\log:z\in\mathbb{C}\setminus{0}\mapsto\ln|z|+i\arg(z)$ where $\ln$ is the real natural logarithm, $|z|$ is the complex modulus of $z$, $\arg$ is the complex argument function such that $\forall{z}\in{({-\infty;0})}:\arg(z)=\pi$ and $\arg(0)=0$. – dbanet Aug 23 '15 at 14:45
  • Functions are very handy to deal with compared to multifunctions. Given a logarithm of a number (any branch), all other branches can be easily enumerated. The same goes for $\arcsin$, $\arccos$, etc. The Lambert $\operatorname{W}$-function is another case, where you can't simply (like $+2\pi i k$) enumerate all values given any particular one. I in particular don't like to have $\log(2)$ not well-defined enough as a real value. – dbanet Aug 23 '15 at 14:57
  • Answer: Practically no. – Marcos Feb 25 '17 at 18:34

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To piggy back off of @Crostul and @dbanet, the answer to the question is that the comic is mathematically accurate. One could even ask does real exponentiation make sense? If you have something like $3^{\pi}$, is that really $$\underbrace{3\cdot 3\cdot\dots \cdot 3}_{\pi \text{ times}}$$How do you multiply something $\pi$ times? Even simpler, how do you even multiply something a rational number of times? The answer is actually not too hard: Rational exponentiation makes some intuitive sense, in that $3^{2/5}=\sqrt[5]{3^2}$, or it is the number such that its 5th power is $3^2$. Then, real exponentiation can be defined to be a limit or supremum of powers of all rational numbers less than the real number. Complex exponentiation requires Euler's formula to make any sort of sense.

For a good reference which, if you've at least seen calculus before should be readable (though maybe difficult), would be Needham's Visual Complex Analysis. It has a great intuitive explanation of Euler's formula.

Moya
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  • Hey! This answers the question I had in the comments section as well. I'd upvote you if I could. Thanks for the explanation (and the link). – Predestination Aug 23 '15 at 13:29
  • Glad to help! Another book which should be readable that has an explanation of real exponentiation is Spivak's Calculus. Look it up if you're interested. – Moya Aug 23 '15 at 13:42
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$a^b=\underbrace{a\cdot a\cdot\ldots\cdot a}_{b~\text{times}}$ is valid only for $b\in\mathbb N_0$. Even for real numbers like $2^{\sqrt{2}}$ you need something different, usually one defines this using exponential and logarithm: $a^b=\exp(b\ln(a))$ for $a>0$.

When looking at $a^z$ with $a\in\mathbb C^*,z\in\mathbb C$ this gets even more complicated as we then have to define the logarithm for complex numbers first and it gets even more "complex" (pun intended) when you consider that there are different branch of logarithms one can use.

Hirshy
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Well, exponentiation isn't just defined for reals. For complex numbers in particular, $e^{i\theta} = \cos(\theta) + i\sin(\theta)$. , and for any positive $b$, $b^z = e^b\ln(z)$.

Eff
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Deusovi
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No, you have to use Euler's formula for imaginary exponents. In fact, as stated here, $\,i^i = e^{-\pi/2}.\,$

Indeed, on complex plane point ${\left(0,i\right)}{}$ corresponds to the angle $\,\theta = \dfrac{\pi}{2}.\,$ Therefore $\,i = e^{\,i\pi/2},\,$ and $$ i^{\,i} = \left(e^{\,i\pi/2}\right)^i = e^{\,i^2\pi/2} = e^{-\pi/2} $$

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Vlad
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